[proofplan] We verify directly that the object and morphism assignments are well-defined and satisfy the two functor axioms. Local smallness ensures that each hom-collection $\operatorname{Hom}_{\mathcal C}(A,X)$ is an object of $\mathbf{Set}$. The identity axiom follows from the identity law in $\mathcal C$, and the composition axiom follows from associativity of composition in $\mathcal C$. [/proofplan] [step:Use local smallness to define the object assignment in $\mathbf{Set}$] Define an assignment $F$ on objects of $\mathcal C$ by \begin{align*} F(X) := \operatorname{Hom}_{\mathcal C}(A,X) \end{align*} for every object $X$ of $\mathcal C$. Since $\mathcal C$ is locally small, $\operatorname{Hom}_{\mathcal C}(A,X)$ is a set for every object $X$. Hence $F(X)$ is an object of $\mathbf{Set}$. [/step] [step:Define the morphism assignment by post-composition] Let $f:X \to Y$ be a morphism in $\mathcal C$. Define \begin{align*} F(f): F(X) &\to F(Y) \\ g &\mapsto f \circ g. \end{align*} This is well-defined because every $g \in F(X)=\operatorname{Hom}_{\mathcal C}(A,X)$ is a morphism $g:A \to X$, so the composite $f \circ g:A \to Y$ is a morphism in $\mathcal C$, hence lies in $F(Y)=\operatorname{Hom}_{\mathcal C}(A,Y)$. [/step] [step:Verify that identities are preserved] Let $X$ be an object of $\mathcal C$, and let $\operatorname{id}_X:X \to X$ denote its identity morphism. For every $g \in F(X)=\operatorname{Hom}_{\mathcal C}(A,X)$, the definition of $F$ on morphisms gives \begin{align*} F(\operatorname{id}_X)(g)=\operatorname{id}_X \circ g. \end{align*} By the identity law in $\mathcal C$, $\operatorname{id}_X \circ g = g$. Therefore $F(\operatorname{id}_X)(g)=g$ for every $g \in F(X)$, so \begin{align*} F(\operatorname{id}_X)=\operatorname{id}_{F(X)}. \end{align*} Thus $F$ preserves identity morphisms. [/step] [step:Verify that composition is preserved] Let $f:X \to Y$ and $h:Y \to Z$ be morphisms in $\mathcal C$. For every $g \in F(X)=\operatorname{Hom}_{\mathcal C}(A,X)$, the definition of $F$ gives \begin{align*} F(h \circ f)(g)=(h \circ f)\circ g. \end{align*} By associativity of composition in $\mathcal C$, \begin{align*} (h \circ f)\circ g = h \circ (f \circ g). \end{align*} Using the definition of $F(f)$ and $F(h)$, this becomes \begin{align*} F(h \circ f)(g) = h \circ F(f)(g) = F(h)(F(f)(g)) = (F(h)\circ F(f))(g). \end{align*} Since this equality holds for every $g \in F(X)$, the functions are equal: \begin{align*} F(h \circ f)=F(h)\circ F(f). \end{align*} Thus $F$ preserves composition. [/step] [step:Conclude that the covariant Hom assignment is a functor] The assignment $F$ sends every object $X$ of $\mathcal C$ to the set $\operatorname{Hom}_{\mathcal C}(A,X)$ and every morphism $f:X \to Y$ to the function $g \mapsto f \circ g$. The preceding steps show that $F$ preserves identities and composition. Therefore $F=\operatorname{Hom}_{\mathcal C}(A,-)$ is a functor $\mathcal C \to \mathbf{Set}$. [/step]